Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r^2 + 3r - 18}{r - 9} \times \dfrac{-7r + 63}{-2r - 12} $
Solution: First factor the quadratic. $q = \dfrac{(r + 6)(r - 3)}{r - 9} \times \dfrac{-7r + 63}{-2r - 12} $ Then factor out any other terms. $q = \dfrac{(r + 6)(r - 3)}{r - 9} \times \dfrac{-7(r - 9)}{-2(r + 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (r + 6)(r - 3) \times -7(r - 9) } { (r - 9) \times -2(r + 6) } $ $q = \dfrac{ -7(r + 6)(r - 3)(r - 9)}{ -2(r - 9)(r + 6)} $ Notice that $(r - 9)$ and $(r + 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -7\cancel{(r + 6)}(r - 3)(r - 9)}{ -2(r - 9)\cancel{(r + 6)}} $ We are dividing by $r + 6$ , so $r + 6 \neq 0$ Therefore, $r \neq -6$ $q = \dfrac{ -7\cancel{(r + 6)}(r - 3)\cancel{(r - 9)}}{ -2\cancel{(r - 9)}\cancel{(r + 6)}} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $q = \dfrac{-7(r - 3)}{-2} $ $q = \dfrac{7(r - 3)}{2} ; \space r \neq -6 ; \space r \neq 9 $